Question

The sum of consecutive natural numbers is 180 more than the average of the numbers. What is the sum of the largest and the smallest of the ten numbers?
A. 45
B. 48
C. 50
D. 40

Answer 1

Hint: Take the terms as \[n,\left( {n + 1} \right),\left( {n + 2} \right),\left( {n + 3} \right){\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}\left( {n + 9} \right)\]. Then, find sum and average and depending upon the question form the equation to find values of smallest and largest terms of the series.

Complete step-by-step answer:
In the question we are given that the sum of ten consecutive natural numbers is 180, more than the average of numbers and from the given information we have to find the sum of the largest and smallest of the ten numbers.
Let's consider smallest number be n, so the other nine numbers will be \[\left( {n + 1} \right),\left( {n + 2} \right),\left( {n + 3} \right),\left( {n + 4} \right),\left( {n + 5} \right),\left( {n + 6} \right),\left( {n + 7} \right),\left( {n + 8} \right),\left( {n + 9} \right)\]
In the question, the term average is given which means sum of all the given terms by total number of terms.
Here, terms are \[n,\left( {n + 1} \right),\left( {n + 2} \right),\left( {n + 3} \right),\left( {n + 4} \right),\left( {n + 5} \right),\left( {n + 6} \right),\left( {n + 7} \right),\left( {n + 8} \right),\left( {n + 9} \right)\]
So, the sum of the term is,
\[n + \left( {n + 1} \right) + \left( {n + 2} \right) + \left( {n + 3} \right) + \left( {n + 4} \right) + \left( {n + 5} \right) + \left( {n + 6} \right) + \left( {n + 7} \right) + \left( {n + 8} \right) + \left( {n + 9} \right)\]
which can be written as\[{\rm{1}}0{\rm{n}} + {\rm{45}}\].
The average of the terms will be sum of the terms by total number of terms or \[\dfrac{{10n + 45}}{{10}} \Rightarrow \left( {n + \dfrac{9}{2}} \right)\]
Hence, according to the question, we can write an equation as:
\[10n + 45 = 180 + n + \dfrac{9}{2}\]
Now, we will subtract 'n' from both the sides of the equation, so, we get,
\[\begin{array}{l}10n + 45 - n = 180 + n + \dfrac{9}{2} - n\\9n + 45 = 180 + \dfrac{9}{2}\\{\rm{Taking\ LCM\ on\ RHS\, we\ get}}\\9n + 45 = \dfrac{{\left( {360 + 9} \right)}}{2}\\9n + 45 = \dfrac{{369}}{2}\end{array}\]
Now, we will subtract by 45 from both the sides, so, we get,
\[\begin{array}{l}9n + 45 - 45 = \dfrac{{369}}{2} - 45\\9n = \dfrac{{369}}{2} - 45{\rm{ }}\\{\rm{Taking\ LCM\ of\ terms\ on\ RHS\, we\ get,}}\\9n = \dfrac{{369 - 90}}{2}\\9n = \dfrac{{279}}{2}\end{array}\]
Dividing by 9 on both sides, we get
\[\dfrac{{9n}}{9} = \dfrac{{279}}{{2 \times 9}}\]
So, the value of n is \[\dfrac{{31}}{2}\]
The smallest term was n which is \[\dfrac{{31}}{2}\] and largest was \[n + 9 \Rightarrow \dfrac{{31}}{2} + 9 \Rightarrow \dfrac{{49}}{2}\]
So, the sum will be \[\dfrac{{31}}{2} + \dfrac{{49}}{2} \Rightarrow \dfrac{{80}}{2} \Rightarrow {\rm{40}}\].
So, the correct option is D.

Note: Instead of taking terms as \[n,\left( {n + 1} \right),\left( {n + 2} \right),\left( {n + 3} \right){\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}\left( {n + 9} \right)\] take an arithmetic progression with first term as n and total number of terms as 10 and common difference as 1. Then, we can use the formula to find sum of n terms as \[Sn = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\], where n is the number of terms, a is first term and d is common difference. So, we will get it as
\[\begin{array}{l}Sn = \dfrac{{10}}{2}\left( {2n + \left( {10 - 1} \right)1} \right)\\Sn = \dfrac{{10}}{2}\left( {2n + 9} \right)\\Sn = 10n + 45\end{array}\]

Now, we can proceed from here as usual.